Question: Simplify the following expression: $y = \dfrac{-9x^2+19x- 2}{x - 2}$
Explanation: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-9)}{(-2)} &=& 18 \\ {a} + {b} &=& &=& {19} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $18$ and add them together. The factors that add up to ${19}$ will be your ${a}$ and ${b}$ When ${a}$ is ${1}$ and ${b}$ is ${18}$ $ \begin{eqnarray} {ab} &=& ({1})({18}) &=& 18 \\ {a} + {b} &=& {1} + {18} &=& 19 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-9}x^2 +{1}x) + ({18}x {-2}) $ Factor out the common factors: $ x(-9x + 1) - 2(-9x + 1)$ Now factor out $(-9x + 1)$ $ (-9x + 1)(x - 2)$ The original expression can therefore be written: $ \dfrac{(-9x + 1)(x - 2)}{x - 2}$ We are dividing by $x - 2$ , so $x - 2 \neq 0$ Therefore, $x \neq 2$ This leaves us with $-9x + 1; x \neq 2$.